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Time of flight is determined only by the vertical component of velocity - it is the time interval between when the projectile was released ($y=y_0$) and when it reaches the ground ($y_t=0$). As the collision is with a vertical wall it acts in the horizontal direction (assuming no friction during the short duration of the collision) and so has no effect on ...

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If the collision is perfectly elastic then there is no change in the kinetic energy during the collision. If there were a change in the kinetic energy during the collision then the flight times would be different. This is analogous to a ball being thrown and encountering no collision.

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You already said if you have $h$ as 1.8 m, then $\dfrac{1}{2} m v^2 = m g h$ implies $v$ is 5.9 m/s. However, you were unhappy becuase this involves energy. So just divide both sides of the equation by $m$. Now you have $\dfrac{1}{2} v^2 = g h$. The solution must still be 5.9 m/s so you get the right answer, this time using only SUVAT.

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I'll risk moderatorial opprobium with a partial answer because you have come so close. You correctly use the SUVAT equation $v^2 = u^2 + 2as$ to find that the velocity of the ball just before it strikes the ground is $v_i = -7$ m/s (using the sign convention that upwards is positive). So far so good. Now you know the ball rises back up to a height of 1.8m, ...

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You're confusing cause and effect. The better equation here is $A={F \over M}$ When a force is applied to an object, the object is accelerated. The object gains kinetic energy from the force, but it doesn't have a property of acceleration. The gun accelerates the bullets with some acceleration for some time period. One gun could accelerate the bullet ...

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There is an underlying assumption for the analysis done by ArtifexR, and to some degree in the OP, that this is a classic physics problem. Most importantly, the collision is treated as an elastic collision which then results in straight-line paths for the two vehicles. It would be very interesting to see the skid marks from this one. At any rate, let's ...

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There a few things to think about in addition to the physics others have mentioned. It's certainly not inconceivable for a larger vehicle to dramatically alter the trajectory of a lighter, slower-moving vehicle. Imagine a bowling ball moving faster than a billiard ball and they collide at right angles. The billiard ball is definitely going to change ...

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There is a transient tension $\Delta T$ in the string that will give rise to a change in momentum of the counterweight, the pan, and the mass on the pan. At the ceiling, the pulley transfers twice this force to the support: The change in momentum of the three components is (with + direction up): $$\Delta p = m v' + m(v-v') - mv' = m(v-v')$$ This change ...

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This is not correct. $\Delta t$ (duration of impulse) is very short indeed, but $F_i$ (force on particle $i$) is very high. It is similar to attempting to compute $\infty\ 0$, which is not zero, but undefined. However, $\int F_i\mathrm{d}t$ obviously is defined, and it is not equal to zero if the velocity of your particle changes during the impulse. The sum ...

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Liquids and gases are both fluids, meaning they flow(duh) and will take the shape of their container. Gases will expand to fill the container, while liquids will not. Solids are not fluids and do not conform to their container's shape. Yes, I know you can squeeze things to fit (e.g. a sponge), but only at the cost of distorting the structure and storing ...

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The forces don't anulate each other because they are not on the same body. The ball creates a force F on the wall. Then, according to the Newton's Third Law, the wall will create a force N with the same module of F but with oposite direction. Pay attention that F is on the wall and N is on the ball, that's why you can't sum them.

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